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Question 28

What pressure will 14.0 g of CO exert in a 3.5 L container at 75°C?

A. 4.1 atm
B. 5.0 atm
C. 6.4 atm
D. 1.1 atm
E. 2.3 atm

Sagot :

Sure, let's solve the problem step-by-step.

To find the pressure exerted by a gas, we can use the Ideal Gas Law, which is formulated as:
[tex]\[ PV = nRT \][/tex]

Where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm).
- [tex]\( V \)[/tex] is the volume in liters (L).
- [tex]\( n \)[/tex] is the number of moles of the gas.
- [tex]\( R \)[/tex] is the ideal gas constant, which is 0.0821 L·atm/(K·mol).
- [tex]\( T \)[/tex] is the temperature in Kelvin (K).

Let's identify and convert all given quantities to the appropriate units.

1. Mass of CO (carbon monoxide): 14.0 g.
- The molar mass of CO is the sum of the atomic masses of carbon (C) and oxygen (O).
[tex]\[ \text{Molar mass of CO} = 12.0 \, \text{g/mol (C)} + 16.0 \, \text{g/mol (O)} = 28.0 \, \text{g/mol} \][/tex]

- Using this, we can calculate the number of moles ([tex]\( n \)[/tex]) of CO.
[tex]\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{14.0 \, \text{g}}{28.0 \, \text{g/mol}} = 0.5 \, \text{mol} \][/tex]

2. Volume (V): 3.5 L.

3. Temperature (T): Given in Celsius (75°C), which needs to be converted to Kelvin (K).
[tex]\[ T = 75 + 273.15 = 348.15 \, \text{K} \][/tex]

4. Ideal gas constant (R): 0.0821 L·atm/(K·mol).

Now substitute all these values into the Ideal Gas Law to solve for the pressure [tex]\( P \)[/tex]:
[tex]\[ P = \frac{nRT}{V} \][/tex]
[tex]\[ P = \frac{(0.5 \, \text{mol}) \times (0.0821 \, \text{L·atm/(K·mol)}) \times (348.15 \, \text{K})}{3.5 \, \text{L}} \][/tex]

After performing the calculations, we get:
[tex]\[ P \approx 4.08 \, \text{atm} \][/tex]

The closest value to our calculated pressure from the given multiple-choice options is:
[tex]\[ \boxed{4.1 \, \text{atm}} \][/tex]